"wouter meeussen" <wouter.meeussen@pandora.be> writes:
One counter-example to spoil things a bit:
What is this a counterexample of?
for n=7, a small sample of (-1,0,1)-matrices whose powers are 'same', shows that not all are pseudo-Antisymmetric (= AntiSymmetric + Diagonal)
What about [ 0 1 ] [ 1 0 ] ? Isn't that a non-PA (-1,0,1)-matrix whose powers are PA (-1,0,1)-matrices? But the (powerlength = divisor of 12) remains valid. That's not true of the permutation matrix [ 0 1 0 0 0 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] [ 0 0 0 0 1 ] [ 1 0 0 0 0 ]. Earlier, you wrote
Consider
the (-1,0,1)-matrices T with properties : Det[T] not zero (invertible), all powers T^k are also invertible (-1,0,1) matrices.
Properties: powerlength of T divides 12, Det[t] is 1 or -1, T is pseudoAntisymmetric,
What are you asserting or conjecturing about those matrices and those properties? By the way, a more standard term for "powerlength" would be "order", or "multiplicative order" if you use the word "order" for some other property. Certainly any matrix of finite multiplicative order must have unit determinant, but I can't see any other likely conjecture. Dan